Divide the following complex numbers. $\dfrac{1+8i}{-2-i}$
Answer: We can divide complex numbers by multiplying both numerator and denominator by the denominator's complex conjugate, which is ${-2+i}$. $ \dfrac{1+8i}{-2-i} = \dfrac{1+8i}{-2-i} \cdot \dfrac{{-2+i}}{{-2+i}} $ We can simplify the denominator using the fact $(a + b) \cdot (a - b) = a^2 - b^2$. $ = \dfrac{(1+8i) \cdot (-2+i)} {(-2)^2 - (-i)^2} $ Evaluate the squares in the denominator and subtract them. $ = \dfrac{(1+8i) \cdot (-2+i)} {(-2)^2 - (-i)^2} $ $ = \dfrac{(1+8i) \cdot (-2+i)} {4 + 1} $ $ = \dfrac{(1+8i) \cdot (-2+i)} {5} $ The denominator now doesn't contain any imaginary unit multiples, so it is a real number. Note that when a complex number, $a + bi$ is multiplied by its conjugate, the product is always $a^2 + b^2$. Now, we can multiply out the two factors in the numerator. $ \dfrac{({1+8i}) \cdot ({-2+i})} {5} $ $ = \dfrac{{1} \cdot {(-2)} + {8} \cdot {(-2) i} + {1} \cdot {1 i} + {8} \cdot {1 i^2}} {5} $ $ = \dfrac{-2 - 16i + 1i + 8 i^2} {5} $ Finally, simplify the fraction. $ \dfrac{-2 - 16i + 1i - 8} {5} = \dfrac{-10 - 15i} {5} = -2-3i $